thermodynamics – chapter - 7- · thermodynamics – chapter - 7- 2 we consider any arbitrary...

Thermodynamics – chapter - 7-

1

CHAPTER 7

ENTROPY

1-The Clausius inequality. The Clausius inequality is a relation

between the temperatures of an arbitrary number of heat reservoirs and the

quantities of heat given up or absorbed by them, when some working substance

is carried through an arbitrary cyclic process in the course of which it

interchanges heat with the reservoirs. For simplicity we shall consider only the

three reservoirs at temperatures oT ,

1T and 2T , shown schematically in

Fig.1, but the argument is readily extended to any number. The rectangle

lettered "system" refers to any device (such as a gas in a cylinder) which is

capable of absorbing and liberating heat with accompanying changes in

volume. The processes taking place in the system are not necessarily reversible

and of course the pipe lines are schematic only.

FIG. 1 Schematic diagram to illustration the Clausius inequality.


2

We consider any arbitrary process in which the system is carried through a

closed cycle, so that its end state is the same as its initial state. Let oQ ,

1Q , and

2Q , represent respectively the quantities of heat interchanged between the system

and the heat reservoirs, and W the net amount of work done by the system. In the

diagram, the system is shown absorbing heat from the reservoirs at temperatures 2T

and 1T , rejecting heat to the reservoir at temperature

oT , and performing

mechanical work, but in the general argument that follows we shall make no

restriction on the directions of these interchanges except that they shall be consistent

with the first and second laws.

To determine whether they are consistent, it is necessary to reduce the general

problem to a process that can be compared either with the Kelvin or Clausius

statements of the second law. Let us introduce as auxiliary equipment two Carnot

engines, operating between pairs of heat reservoirs as shown. (In the diagram, the

engines are actually refrigerators. We shall use the generic term, "engine," since in

the general case some might be operated as engines and others as refrigerators.)

Engine A supplies to the reservoir at temperature 1T a quantity of heat

AQ1 equal

to that given up by this reservoir in the original cyclic process. This engine withdraws

heat AQ0

from the reservoir at temperature oT and work

AW is supplied to it. By

this process we eliminate any outstanding change in the reservoir at temperature 1T .

Carnot engine B supplies heat BQ2

to the reservoir at temperature 2T , in

amount just equal to that given up to the system. It also withdraws heat BQ0 from

the reservoir at temperature oT and work

BW is supplied to it. Thus there has been

no outstanding change in the reservoir at temperature 2T . Since the original process

through which the system was carried was cyclic, everything is now as it was at the

start, except for the reservoir, at temperature oT and the mechanical system that


3

supplied or absorbed work. 'Unless oQ happens to be equal to the sum of

AQ0 and

BQ0, the reservoir at temperature

oT will have gained or lost some heat and unless

W and BA WW are equal, the mechanical system has gained or lost work. Of

course, we know from the first law that the net flow of heat from or to the heat

reservoir must equal the net work clone on or by the mechanical system and, as far as

the first law is concerned, it is a matter of indifference whether the heat reservoir

loses or gains heat, as long as the mechanical system gains or loses the same quantity

of work. It is not, however, a matter of indifference from the point of view of the

second law, since the second law (Kelvin statement) obviously would be violated if

the heat reservoir lost heat and the mechanical system gained an equal amount of

work. Therefore the heat reservoir must have gained heat and the mechanical system

given up work, except in the special case where both have neither lost nor gained.

That is, the heat oQ must be greater than the heat BA QQ 00 or, in the limiting

case, just equal to it.

Before putting these conclusions in analytic form, it is necessary to adopt a

convention of sign for the SQ' . We shall write our equations from the point of view

of the system. That is, a quantity of heatQ is considered positive if heat is given to

the system and negative if the system gives up heat. Since 2Q , for example, is at the

same time heat given to the system and heat given up by a reservoir, it follows that

heat given up by a reservoir is Positive, heat given to a reservoir is negative.

We can now write the following equations:

00

0

1

1 T

Q

T

Q AA, 011 QQ A , -------------- ( 1 )

00

0

2

2 T

Q

T

Q BB, 022 QQ B , -------------- ( 2 )


4

In the diagram as drawn, 1Q and

2Q are positive, oQ is negative,

AQ1 is

negative, AQ0

is positive, BQ2

is negative, BQ0

is positive. However, in setting

up the general equations, we write these all as positive quantities just as one writes

0,0 yx FF for forces in equilibrium at a point, or , 0i in

Kirchhoff's point rule.

From Eqs. (1) we have

1

1

T

QTQ ooA

-------------- ( 3 )

and from Eqs. (2),

2

2

T

QTQ ooB

-------------- ( 4 )

The net quantity of heat given up by the reservoir at temperature oT is

oBoAo QQQ . But the second law requires that the reservoir can only receive

heat, not give it up. Therefore the sum oBoAo QQQ must be a negative

quantity or, in the limiting case, zero.

0 oBoAo QQQ .

Inserting the expressions for AQ0

and BQ0

from the preceding equations, we get

02

1

1

o

oooT

QT

T

QTQ


5

After dividing both sides of the inequality by oT , this reduces to the

symmetrical form

02

2

1

1 T

Q

T

Q

T

Q

o

o

It should be evident that by making use of a sufficiently large number of

Carnot engines, a process in which the system in Fig. 1 interchanged heat with any

number of reservoirs could be reduced to an energy interchange between a single heat

reservoir and a work reservoir. We would

then find

. -------------- ( 5 )

By extension, when the number of reservoirs becomes infinitely large and the

system may exchange only infinitesimal amounts of heat with each, the finite sum

becomes an integral and

-------------- ( 6 )

Either Eq. (5) or (6) expresses the Clausius inequality, a relation between the

quantities of heat absorbed or liberated by a number of heat reservoirs and the

temperatures of those reservoirs when a working substance is carried through a

cyclic process of any sort, reversible or irreversible. The quantities Q or Qd / are

considered positive when heat is given up by a reservoir. The inequality makes no

statements about the system that was carried through the original process, which it

obviously cannot do, since at no point did the temperature of the system enter the

argument. The only temperatures considered are those of the reservoirs.

We give three examples of the Clausius inequality. First, consider the flow of

heat by conduction from a reservoir at a temperature 2T to a reservoir at a lower


6

temperature 1T . This is a very simple case in which no "system" at all is involved,

other than the material which conducts heat from one reservoir to the other. The

quantities of heat 2Q and

1Q are necessarily equal in magnitude, but 2Q is

positive while 1Q is negative. As a numerical example, let T2 = 400 K, T1 = 200 K,

Q2 = 800 joules, Q1 = - 800 joules, then

.deg/242200

800

400

800Joules

T

Q

and we see that T

Q is actually less than zero.

As a second example, consider a heat engine operating between reservoirs at

400 K and 200 K. The efficiency of a Carnot engine operating between these

temperatures would be 50%. Let, us assume that our engine is less efficient than a

Carnot engine, with an efficiency of only 25% (we have shown that it cannot be in

more efficient). Assume that the engine takes in 800 joules from the high

temperature reservoir. It then converts 25% of this, or 200 joules, into work and

rejects 600 joules to the low temperature reservoir. The engine then becomes the

system in our general argument. We have

.deg/132200

600

400

800Joules

T

Q

and again T

Q is less than zero.

Finally, consider a Carnot engine operating between reservoirs at 400 K and

200 K. Its efficiency is 50 % and if it takes in 800 joules it rejects 400 joules. Hence


7

022200

400

400

800

T

Q

and in this case the equality sign holds, rather than the inequality.

2-Entropy. In deriving the Clausius inequality, no restrictions were placed on

the reversibility or irreversibility of the cycle through which the system was carried.

Let us now assume the cycle to be reversible, and let the system traverse it first in one

direction and then in the opposite direction. Let 1

/Qd represent the heat flowing

into the system at any point in the first cycle, and 2

/Qd the heat flowing into it at

the same point in the second cycle. Since all processes in the second cycle are the

reverse of those in the first, then

2/

1/ dQQd ----------------- ( 7 )

If the cycle is reversible, the temperature of the - while it, is exchanging heat with

any reservoir is the same as the temperature of that reservoir. We can therefore write

the Clausius inequality for the two cycles as

----------------- ( 8 )

The symbol ( ) means that the integration is carried around a cyclic path. We

could not use this symbol in Eq. (6) because this equation applied to the reservoirs

which were not carried through cyclic processes. Combining Eqs. (7) and (8), we get

But the only way in which both of these relations can be true is if the equality

sign holds, and not the inequality. We therefore have the very important result that

when a system is carried around a reversible cycle and the heat dT2 added to it at

every point is divided by its temperature at that point, the sum of all such quotients

is zero.


8

----------------- ( 9 )

Consider now an arbitrary reversible cycle such as that represented by the closed

curve in Fig. 2. Points 1 and 2 are any two points on the curve. The integral around

the closed curve can be written as the sum of two integrals, one along path I from 1

to 2, the second along path II from 2 back to 1.

Fig. 2

----------- ( 10 )

If the path II from 2 to 1 were traversed in the opposite direction, we would have, since the cycle is reversible,


9

----------- ( 11 )

Combining the result with eq (10 ), we get

----------- ( 12 )

That is, the integral is the same along the two reversible paths. Also, since the

original cycle was entirely arbitrary, it follows that the integral is the same along all

reversible paths from 1 to 2. In mathematical terms, the quantity

T

Qd /

is an exact

differential of some function S of the state of the system, and may be represented

by dS . Then

---------- ( 13)

Since the integral of an exact differential along any path is equal to the difference

between the values of the function at the end points of the path. The quantity S is

called the entropy of the system, and Eq. (13) states that the change in entropy of a

system between any two equilibrium states is found by taking the system along any

reversible path connecting the states, dividing the heat added to the system at each

point of the path by the temperature of the system, and summing the quotients thus

obtained.


10

Thus we see that while Qd / is not an exact differential, there are two ways in

which an exact differential involving Qd / can be obtained. One is to subtract

from Qd / the inexact differential Wd / . The sum is the exact differential

dU and the statement of this fact constitutes the first law of thermodynamics.

------------- ( 14 )

The second method is to divide Qd / by the temperature T . The quotient is the

exact differential dS and the statement of this fact constitutes the second law of

thermodynamics.

------------- ( 15 )

In the mks system, entropy is expressed in joules per degree Kelvin. In

engineering, the Btu per degree Rankine is used. Another common unit is the calorie

per degree Kelvin.

The following points should be noted.

(a) The entropy of a system is defined for equilibrium states only.

(b) Only changes in entropy or entropy differences can be computed from Eq.

(13). In many practical problems, such as the design of steam engines, we are

concerned only with differences in entropy. For convenience, the entropy of a

substance may be assumed zero in some convenient reference state and a numerical

value can then he assigned to the entropy in any other state. Thus in computing

steam tables, the entropy of water is assumed zero when it is in the liquid phase at

0° C and 1 atmosphere pressure, and a value is listed for the entropy over a wide

range of temperatures and pressures, in the vapor as well as in the liquid phase.

(c) The entropy of a system in an equilibrium state is a function of the state of

the system only and is independent of its past history. The entropy can therefore be

expressed as a function of the thermodynamic coordinates of the system, such as the

pressure and temperature or pressure and volume.


11

(d) Changes in entropy can be computed from Eq. (13) for reversible

processes only. However, any reversible process between the same two end states

can be used, since the change in entropy is the same for all reversible processes

between the same two end states.

(e) To compute the change in entropy of a system when it goes from one

equilibrium state to another equilibrium state by an irreversible process, several

methods can be used:

(1) We can devise a reversible process connecting the same two end

states and use Eq. (13).

(2) If a table such as a steam table has already been prepared, tabulating

the entropy in a large number of states, we can read the entropies at the end states

from the table and subtract one from the other.

(3) If the equation for the entropy as a function of the thermodynamic

coordinates of the system is known, the entropies at the end states can be calculated

from this equation and one subtracted from the other.

Methods (2) and (3) can also be applied to reversible processes.

Entropy is an extensive property of a system, proportional to the mass of the

system or to the number of moles. We shall make frequent use of the specific

entropy, represented by S and defined by

,m

Ss

or by .

n

Ss

3-Calculation of changes in entropy. We now give a few examples of the

methods used to compute changes in entropy. To begin with, it is obvious that when a

system undergoes a reversible adiabatic process there is no change in its entropy,

since by definition the heat absorbed in such a process is zero. A reversible adiabatic

process therefore proceeds at constant entropy and may be described as isentropic.

The entropy is not constant in an irreversible adiabatic process and we shall return to

this point later.

The simplest process involving a change in entropy is a reversible process at

constant temperature. For such a process the constant temperature T may be taken

outside the integral sign and the general equation.


12

------------------- ( 16 )

That is, the change in (specific) entropy of a system in a reversible isothermal

process equals the heat absorbed by the system, per unit mass or per mole, divided by

the Kelvin temperature of the system.

A common example of a reversible isothermal process is a change in phase at

constant pressure, during which, as we have seen, the temperature remains constant

also. To carry out the change reversibly, the system is brought into contact with a

heat reservoir at a temperature infinitesimally greater than the equilibrium

temperature at the given pressure. The change in phase then proceeds very slowly and

at all stages of the process the system is essentially at the equilibrium temperature T .

The heat qT absorbed by the system, per mole or per unit mass, equals the heat of

transformation l and the change in (specific) entropy is merely

T

lss 21

------------------- ( 17 )

In most processes, a reversible absorption of heat is accompanied by a change

in temperature, and calculation of the corresponding entropy change requires an

evaluation of the integral of d'q/T. It is therefor necessary to express d`q in terms T

or vice versa, or to express both in terms of a single variable. If the process takes

place at constant volume for example, and if changes in phase are excluded, then

dTcqd /


13

------------------ ( 18 )

If the process is at constant pressure, pp dTcqd /

………………… ( 19 )

A system originally at the temperature T1 could be raised to the temperature T2

by bringing it in contact with a heat reservoir at temperature T2. Such a process

would be highly irreversible, since there would be large temperature differences

between parts of the system, and between the system and the reservoir. The same

end result can be attained reversibly with the help of a large number of heat

reservoirs, as described in ch.1. During each stage of a process carried out in this

way, the temperature of the system increases by dT and each stage is essentially

reversible and isothermal. The increase in entropy in the system is the same

weather the process is reversible or not, since the change in entropy depends only

on the end points. Where, then, is there any essential difference between the

reversible and irreversible processes? We shall answer this question in Sec. 4.

As a numerical example of the processes just described, let us compute the

increase in the specific entropy of water when it is heated at constant atmospheric

pressure from a temperature of 200 K (ice) to a temperature of 400 K (superheated

steam). The process is represented by the line abcdef in Fig. 3(a). For simplicity, we

shall ignore variations in specific heat capacity with temperature and assume that

cp (ice) = 2.09 X 103 joules/kgm-deg (= 0.50 cal/gm-deg),

cp (water) = 4.18 X 103 joules/kgm-deg (= 1.00 cal/gm-deg),

cp (steam) = 2.09 X 103 joules/kgm-deg (= 0.50 cal/gm-deg).

l12(273° K) = 3.34 X 105 joules/kgm (= 80 cal/gm),

L23 (373° K) = 22.6 X 105 joules/kgm (= 540 cal/gm).


14

FIG. 3. Entropy changes in an isobaric process.

The first stage of the process consists of heating the ice from 200 K to its

melting temperature of 273K. During this process the increase in entropy is

.deg/651200

273ln1009.2ln 3 KgmJoulesxx

T

Tcss

a

bpab

The increase in entropy when the ice is melted is

.deg/1230273

1034.3 512 KgmJoules

x

T

lss bc

The increase in entropy of the liquid water when heated from 273 K to 373 K is


15

.deg/1310273


T

Tcss

c

dpcd

In the process of vaporizing the water at 373 K the entropy increase is

.deg/6060373

106.22 523 KgmJoules

x

T

lss de

Finally, the increase in entropy on heating the vapor to 400 K is

.deg/146373


T

Tcss

e

fpef

In engineering work, the entropy of water is arbitrarily set equal to zero in the

liquid phase at 273 K and atmospheric pressure. Relative to this reference state, then,

the specific entropy of the saturated liquid at 373 K and 1 atm is 1310 joules/kgm-

deg, that of saturated vapor at the same temperature and pressure is 1310 + 6060=

7370 joules/kgma-deg, and that of superheated steam at 400 K and 1 atm is 7370 +

146 = 7520 joules/ kgm-deg. The entropy of the saturated solid at 273 K and 1 atm is

-1230 joules/kgrn-deg, and that of the solid at 200 K and 1 atm is -1230 – 651= -1880

joules/kgm-deg. The figures above are only approximate, since we have neglected

variations in the specific heat capacity with temperature.

The entropy changes in the process are shown in the graph of Fig. 3-b, where

the points lettered a, b, c, d, e, f corresponds to those in Fig. 3-a.

4- Entropy changes in irreversible processes. We next the change in

entropy of a system in an irreversible process. Suppose two bodies at different

temperatures are brought into contact with a rigid adiabatic enclosure and allowed to

come to thermal equilibrium. The equilibrium state will be attained spontaneously

without the help or work reservoirs. If we consider the two bodies together to institute


16

a system, no heat is absorbed or given up by the system and (

T

Qd / ) is zero. It

does not follow that the change in entropy of the system is zero, however, because the

change in entropy of a system is given by (

T

Qd /) for reversible processes only

and the flow of heat through a finite temperature difference is an irreversible process.

Let us use method (e) (1) in Sec. 2 to find the change in entropy. That is, we

devise a reversible process leading from the same initial to the same final state. To be

specific, consider the process of mixing 1 kgm of water at 373 K ( 100° C) with 1.

kgm of water at 273 K, (0° C). If small variations in specific heat capacity are

neglected, elementary calorimetry shows that the final equilibrium temperature is

323 K (50° C). physical mixing of the hot and cold water is, of course, not essential.

The same end state of 2 kgm of water at 323° K will be reached if the hot and cold

water are allowed to exchange heat by conduction or radiation.

To reach this end state reversibly, we can make use of an infinite series of

heat reservoirs at temperatures ranging from 273 K to 373 K. Heat the water

originally at 273 K reversibly, using the reservoirs between 273 K and 323 K, and

cool the water originally at 373 K in a similar manner. We then have 2 kgm of water

at 323 K and the entire process has been reversible.

The increase in entropy of 1 kgm of water when heated from 273 K to 323K is

.deg/704273

323ln1018.41 3

12 Joulesxxxss

The increase in entropy of 1 kg of water when cooled from 373 K to 323K is

.deg/603373

323ln1018.41 3/

1/

2 Joulesxxxss


17

The negative sign means that the entropy has decreased.

The net change in entropy of the water is

,deg/101603704 Joules

and the entropy of the water has increased by 101 joules/deg. in the reversible

process. But the change in entropy of the water is the same in the irreversible

process of mixing, since the end states are the same in both processes, and the

problem is solved. We must note carefully an important difference between the process in which the hot and cold water are mixed directly and that in which they are brought to their final temperature by reversible heat exchange with a series of reservoirs. In the first

process, no bodies other than the two masses of water are involved, and the net increase in entropy of the water is also the net increase in entropy of the universe. In the second process the heat reservoirs are involved as well is the water, and the change in entropy of the reservoirs at each stage is equal and opposite to that of the water. Hence, although the entropy of the water increases in the reversible process, the entropy of the reservoirs decreases by exactly the same amount and the net change in entropy of the universe is zero.

We can now gain a further insight into the significance of the terms reversible

and irreversible. According to the definition previously given, that a reversible

process is a succession of equilibrium states, the mixing of the hot and cold water is

an irreversible process. The fact that it is irreversible in the thermodynamic sense

does not imply, of course, that the 2 kgm of water at 323 K cannot be restored to its

original state of 1 kgm at 373 K and 1 kgm at 273 K. But let us see what happens

when we do restore the original state. Let us heat. 1 kgm from 323 K to 373 K,

using the heat reservoirs between 323 K and 373 K, and cool 1 kgm from 323 K to

273 K. using the reservoirs between 323 K and 273 K. By the same methods as

used in Sec. 3, we find that the entropy of the water heated from 323 K to 373 K

has increased by 603 joules/deg, while that of the reservoirs from which it absorbed

heat has decreased by the same amount. The entropy of the water cooled from 323

K to 273 K has decreased by 704 joules/deg and that of the reservoirs has increased

by 704 joules/deg. Therefore the net decrease in entropy of the water has been 101

joules/deg. and the net increase in entropy of the reservoirs has been 101

joules/deg. The entropy of the water has decreased by the same amount that it increased in

the mixing process, so that its entropy is the same as it was originally. However, the entropy of the reservoirs has increased, and by an amount just equal to the increase


18

in the entropy water in the mixing process. Thus while any outstanding change in the entropy of the water has been eliminated, the entropy increase in the original irreversible mixing process has been passed along to the reservoirs. Any other process we may device to restore the reservoirs to their initial condition will be found to end with an entropy increase of some other system or systems, at least as great as that in the original irreversible process. Hence, while a finite, system can always be restored to its original state after an irreversible process, the increase in entropy associated with the process can never be wiped out. At most, it can be passed on from one system to another. This is the true significance of the term irreversible. We

shall give in Sec. 5 a general proof of the fact that the entropy of an isolated

system increases in any irreversible process. Fig. 4, which should be studied carefully, illustrates the processes described above.

FIG. 4


19

The reader may well ask at this point, "Who cares whether a process is

reversible or not? Suppose the entropy of the universe has been irretrievably

increased in an irreversible process? What of it? There has been no loss of energy in

the process. In what significant way have things been changed?"

One answer to the question above is that the entropy change in a process is the

criterion we sought at the beginning of the chapter, namely, it determines the

direction in which a given process, consistent with the first law, will go. The physical

chemist is chiefly concerned with this aspect of entropy. Will two substances react

chemically or will they not? If the reaction would result in a decrease in entropy, the

reaction is impossible. However, while the entropy might decrease if the reaction

were to take place at one temperature and pressure, it is possible that it could increase

at other values of temperature and pressure. Hence a knowledge of the entropies of

substances as functions of temperature and pressure is all-important in determining

the possibilities of chemical reactions.

The mechanical engineer is interested in reversibility and entropy changes for a

somewhat different reason. From his point of view, something has been "lost" when

an irreversible process takes place in a steam engine or turbine. What is lost,

however, is not energy, but opportunity the opportunity to convert internal energy to

mechanical energy. Since the internal energy of the working substance in a heat

engine is usually replenished by a flow of heat into it, we often say that what is lost is

an opportunity to convert heat to mechanical work. Let us illustrate by the example of

mixing hot and cold water. The internal energy of the system is the same before and

after mixing, but at the end of the process we have a single reservoir (considering the

water now as a reservoir) all at one temperature, whereas at the beginning we had two

reservoirs at different temperatures. It is impossible to withdraw heat from a single

reservoir and operate a cyclic engine, whereas we could have operated an engine

between the original hot and cold reservoirs, withdrawing heat from one rejecting

heat to the other, and diverting a part of the heat to produce' mechanical work. Once

the reservoirs have come to the same temperature , this opportunity is irretrievably

lost. Thus any irreversible process in a heat engine reduces its efficiency, that is, it

reduces the amount of mechanical work that can be abstracted from a given amount

of heat absorbed by the working substance.


20

5-The principle of the increase of entropy. All actual processes are

irreversible. They take place at a finite rate, with finite differences of temperature and

pressure between parts of a system or between a system and its surroundings. We

now show that it is a necessary consequence of the second law that the entropy of an

isolated system increase in every natural (i.e., irreversible) process.

In mechanics, one of the reasons that justify the introduction of the concepts of

energy, momentum, and angular momentum is that they obey a conservation

principle. Entropy is not conserved, however, except in reversible processes, and this

unfamiliar property, or lack of property, of the entropy function is one reason why

such an aura of mystery usually surrounds the concept of entropy. When a beaker of

hot water is mixed with a beaker of cold water, the heat lost by the hot water equals

the heat gained by the cold water. "Heat" is conserved in this process or, more

generally, energy is conserved. On the other hand, while the entropy of the hot water

decreases in the mixing process and the entropy of the cold water increases, the

decrease in entropy is not equal to the increase, and the total entropy of the system is

greater at the end of the process than it was at the beginning. Where did this

additional entropy come from? The answer is that the additional entropy was created

in the process of mixing the hot and the cold water. Furthermore, once entropy has

been created, it can never be destroyed. The universe must forever bear this

additional burden of entropy. "Energy can neither be created nor destroyed," says

the first law of thermodynamics. "Entropy cannot be destroyed," says the second

law, "but it can be created."

In Fig. 5, the crosshatching represents a natural (and hence irreversible)

process taking place in an isolated system. As a result of this process, the system

moves from an equilibrium state represented by point 1 to another equilibrium state

represented by point 2. The continuous line represents a reversible process, involving

interchanges of energy with heat and work reservoirs, by which the system is returned

from state 2 to state 1. Taken together, processes I and II constitute a cycle. The

cycle as a whole is irreversible, since part I is irreversible. Hence, from the Clausius

inequality,


21

fig 5, A system undergoes an irreversible process from state (1) to state (2), and is

returned by a reversible process from state (2) to state (1).

or, writing the integral as the sum 'of two integrals,

But the first integral is zero, since the system was isolated in the irreversible

process and could not receive or give out heat. This integral, however, is not equal to

12 SS , since only for reversible processes is pp dTcqd /. The second

integral, since path II is reversible, is 21 SS It follows that

021 SS , or 12 SS

That is, the entropy of the system in state 2 is greater than in state 1. Since the

original process was arbitrary, we conclude that the entropy of an isolated system

increases in every natural (i.e., irreversible) process.


22

Note that the statement above is restricted to isolated systems and that the

entropy refers to the total entropy of the system. When natural. Processes take place

in an isolated system, the entropy of parts of the system may decrease and that of

other parts may increase. The increases, however, are always greater than the

decreases. The entropy of a non- isolated system may either increase or decrease,

depending on whether - heat is added to or removed from it or whether irreversible

processes take ,place within it. Hence, the discussing increases and decreases in

entropy, it is very important that the system under consideration shall be clearly

defined.


23

Questions

Q.(1) A 20-ohm resistor carrying a constant current of 10 amp is kept at a constant

temperature of 27° C by a stream of cooling water. in a time interval of 1 second,

(a) What is the change in entropy of the resistor?

(b) What is the change in entropy of the universe?

Q.(2) A thermally insulated 20-ohm resistor carries a current of 10 amp for 1 sec.

The initial temperature of the resistor is 10° C, its mass is 5 gm, and its specific heat

capacity is 850 joules/kgm-deg.

(a) What is the change in entropy of the resistor?

(b) What is the change in entropy of the universe?

Q.(3) One kilogram of water is heated by an electric heating coil from 20° C to 80°

C. Compute the change in entropy of (a) the water, (b) the universe.

Q.(4) 1 Kgm . Of water at a temperature of 280 K is mixed 2 kgm of water at a

temperature of 310 K in a thermally insulated vessel. Find the change in entropy of

the universe.

Q.(5) A mass m of a liquid at a temperature T1 is mixed with an equal mass of the

same liquid at a temperature T2. The system is thermally insulated. Show that the

entropy change of the universe is

21

21

21

2ln2

TT

TT

mcp


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Q.(6) a// One kilogram of water at 0 C° is brought into contact with a large heat

reservoir at 100o C. When the water has reached 100C°. what has been the change in

entropy of the water? Of the heat reservoir? Of the universe ?.

b// if the water had been heated from 0o C to 100o C by first bringing

it in contact with a reservoir at 50o C and then with a reservoir at 100o C, what would

have been the change in entropy of the universe?.

Q.(7) The value of Cp for a certain substance can be represented by Cp = a + bT

a// Find the heat absorbed and the increase in entropy of a mass m of the substance

when its temperature is increased at constant pressure from T1, to T2.

b// Find the increase in the molar specific entropy of copper, when the temperature is

increased at constant pressure from 300K to 1200K.

Q.(8) 10Kgm. of water at a temperature of 20o C is converted to superheated steam

at 250o C and at constant atmospheric pressure compute the change in entropy of the

water. Cp(L) = 4.18x103 J/kg.deg.. l23( at 100o C) = 22.6 x 105 J/kg.deg.

C(vapor) =1670+ 0.494T + 1.86x105T-2 J/kg.deg.

Q.(9) Find the increase in entropy when 20 gm of ice at 0o C melt into water at the

same temp. Given l12= 80 cal/gm

Q.(10) Find the change in entropy when 5 grams of water boils under a pressure of

174.39 Cm given boiling point of water under a pressure of 174.39 Cm = 125o C

and latent heat of steam at the same temperature = 518.7 cal/gm?.


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Q.(11) Calculate the increase in entropy when one gm. of ice at ( -10o C ) is

converted into steam at 100o C superheated of ice = 0.5 latent heat of ice = 80 cal/gm,

latent heat of the steam = 540cal/gm.?.

Q.(12) 10gm of water at a temperature of 20o C is converted into ice (-10o C) at

constant atmospheric pressure assume that superheated of ice at constant pressure to

be 0.5 cal./gm. Find the total change in entropy of the system?.

Q.(13) 10gm of steam at 100o C are blown into 90gm of water at 0o C contained in

a Calorimeter of water equivalent 10gms. The whole of steam is condensed.

Calculate the increase in entropy of the system, latent heat of steam = 540cal/gm.

Q.(14) A volume of a gas expands isothermally to twice it is initial volume.

Calculate the change in entropy in terms of the gas constant.

Q.(15) Calculate the change in entropy when 10gm of ice at 0o C is converted into

water at the same temperature.

Q.(16) Calculate the change in entropy when 5gm of water at 100o C is converted

into steam at the same temperature.

Q.(17) Calculate the increase in entropy when 1gm in of ice at (-10o C) is converted

into steam at 100o C, specific heat of ice = 0.5 latent heat of ice = 80cal/gm. latent

heat of steam =540cal/gm.


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Q.(18) 1gm of molecule of a gas expands isothermally to 4times its volume.

Calculate the change in its entropy in terms of the gas constant.

Q.(19) 50gm of water at 0o C is mixed with an equal mass of water at 83o C.

Calculate the resultant increase in entropy.

Q.(20) Calculate the change in entropy when 50gms. of water at 15o C is mixed

with 80gm of water at 40o C Cp(water) = 1.


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thermodynamics – chapter - 7- · thermodynamics – chapter - 7- 2 we consider any arbitrary...

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